PLEASE HELP!!!!!!!! what is log base 3 six square root 243 (where 6 is inside the "v" of the square root"

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[\log _{3} \sqrt[6]{243}\]

- anonymous

\[\log_{3}\sqrt[6]{243} \] like this?

- anonymous

yes.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

alright lemme see what i can do

- anonymous

okay, can u help me with a few others as well? i am really not understanding logs :(

- anonymous

no problem. You know the rule of logs? Where logANS / logBASE = expo?

- anonymous

?

- anonymous

so in this case log [\sqrt[6]{243}\] / log3
= 5/6

- anonymous

i don't really get it..can you draw it out?

- anonymous

|dw:1327550139579:dw|

- anonymous

iam not allowed to use a calculator for these so i have to get it in the exponent form

- anonymous

\[\frac{1}{6}\log_3(243)\] is a start

- anonymous

and since
\[243=3^5\] you get
\[\frac{1}{6}\log_3(3^5)=\frac{5}{6}\]

- anonymous

|dw:1327550493647:dw|

- anonymous

was i on the right rack?

- anonymous

Yes. Now use logs to make the expo turn into a coefficient

- anonymous

how do i do that ...?

- anonymous

log3^y = log243^(1/6)

- anonymous

Now you can put the y in front of the log as well as the 1/6
so it becomes
ylog3 = (1/6)[log243]

- anonymous

i am so confused :'''(

- anonymous

shall we go off satelite73's method? We keep it as a log form.

- anonymous

i have to do this non calc

- anonymous

So from the \[\sqrt[6]{}\] we can turn it into an exponent as 1/6
so\[\log_{3}243^{1/6}\]

- anonymous

i get that now, so im using satelites method as that does not require a calc.

- anonymous

really your life would be a lot easier if you put the 1/6 out front

- anonymous

i said that too

- anonymous

\[\log(x^n)=n\log(x)\] so
\[\log(\sqrt[6]{x})=\frac{1}{6}\log(x)\]

- anonymous

i am really not understanding anything. i just can't do this :(

- anonymous

The reason we use logrithims for unknown exponential equations is to turn it from an exponent to a coefficient. It then becomes an algebraic equation. So in this situation we have \[\log_{3}\sqrt[6]{243} \]
We turn the \[\sqrt[6]{243}\]
into \[243^{1/6}\]
So right now we have
\[\log_{3} 243^{1/6}\]
now since we have the log, we are allowed to drop the 1/6 down from the exponent level to become the leading coefficient thus making the equation
\[1/6\log_{3}243 \]
From here we can change the number 243 into a number with a base of 3.
We know that 3^5 = 243
therefore we can write
\[1/6\log_{3}(3^5) \]
And knowing that when a base and the number above it are the same number u can cancel out the log and it's base, leaving you with just the 5. So you are to be left with 1/6 * 5
this is known to be 5/6

- anonymous

dang. thank you so much! alb to help me with anymore?

- anonymous

Yessir! glad to help :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.